Đáp án:
a)
$m_{Cu}=28,8(g)$
$m_{CuO}=1,2(g)$
b)
$\%Cu=96\%$
$\%CuO=4\%$
Giải thích các bước giải:
Phản ứng xảy ra:
\(3Cu + 8HN{O_3}\xrightarrow{{}}3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\)
\(CuO + 2HN{O_3}\xrightarrow{{}}Cu{(N{O_3})_2} + {H_2}O\)
Ta có:
\({n_{NO}} = \dfrac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)
\( \to {n_{Cu}} = \dfrac{3}{2}{n_{NO}} = 0,45{\text{ mol}}\)
\( \to {m_{Cu}} = 0,45.64 = 28,8{\text{ gam}} \\\to {{\text{m}}_{Cu}} = 30 - 28,8 = 1,2{\text{ gam}}\)
\( \to \% {m_{Cu}} = \dfrac{{28,8}}{{30}}.100\% = 96\% \\\to \% {m_{CuO}} = 4\% \)